Subject description:
In Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general public. Every time a trial is set to begin, a jury has to be selected, which is done as follows. First, several people are drawn randomly from the public. For each person in this pool, defence and prosecution assign a grade from 0 to 20 indicating their preference for this person. 0 means total dislike, 20 on the other hand means that this person is considered ideally suited for the jury.
Based on the grades of the two parties, the judge selects the jury. In order to ensure a fair trial, the tendencies of the jury to favour either defence or prosecution should be as balanced as possible. The jury therefore has to be chosen in a way that is satisfactory to both parties.
We will now make this more precise: given a pool of n potential jurors and two values di (the defence's value) and pi (the prosecution's value) for each potential juror i, you are to select a jury of m persons. If J is a subset of {1,..., n} with m elements, then D(J ) = sum(dk) k belong to J
and P(J) = sum(pk) k belong to J are the total values of this jury for defence and prosecution.
For an optimal jury J , the value |D(J) - P(J)| must be minimal. If there are several jurys with minimal |D(J) - P(J)|, one which maximizes D(J) + P(J) should be selected since the jury should be as ideal as possible for both parties.
You are to write a program that implements this jury selection process and chooses an optimal jury given a set of candidates.
Input:
The input file contains several jury selection rounds. Each round starts with a line containing two integers n and m. n is the number of candidates and m the number of jury members.
These values will satisfy 1<=n<=200, 1<=m<=20 and of course m<=n. The following n lines contain the two integers pi and di for i = 1,...,n. A blank line separates each round from the next.
The file ends with a round that has n = m = 0.
Output:
For each round output a line containing the number of the jury selection round ('Jury #1', 'Jury #2', etc.).
On the next line print the values D(J ) and P (J ) of your jury as shown below and on another line print the numbers of the m chosen candidates in ascending order. Output a blank before each individual candidate number.
Output an empty line after each test case.
answer:
This is a 0/1 knapsack problem, each candidate has the election and not vote for two options.
Provided $ f [i, j, k] $ $ I represents is selected to the second individual time $, $ J $ has been selected for the individual, and the prosecution subtracting out the defense out when the value of $ k $ , out of the defense and prosecution combined score the maximum (absolute value of the minimum required because the subject when the maximum score).
It is not difficult to draw the following dynamic transfer equation:
$f[i,j,k]=max(f[i-1,j,k],f[i-1,j-1,k-(a[i]-b[i])])$
Wherein the array represents a fraction of the prosecution, b represents an array of points in the defense.
Careful observation can be found, $ i $ this dimension can be omitted, which means we need $ j $ with a dimension To reverse circulation (so we can ensure that every candidate elected only once).
We also claim the subject after the output path, we use $ d [i, j, k ] $ $ I $ denotes the cycle time of the final dimension of which is selected from the individual to determine the array d, we can along the array d the transfer path of the recording, to give a recursive scheme.
Specifically provided is $ last = d [i, j , k] $, continuously from the state $ (i, j, k) $ recursively to $ (last-1, j- 1, k- (a [last] -b [ last])) $, up to $ j = 0 $.
The following code (AC code below and can not, because of some strange questions runtime errors, but I went online to find a can of AC code and run random test data, the results should be no problem):
#include<iostream> #include<algorithm> #include<cstring> using namespace std; int f[30][1000],d[205][30][1000],a[205],b[205],p[900]; int pos=900,num,cnt,u,D,P,t; void dfs(int i,int j,int k){ p[++cnt]=d[i][j][k]; int last=d[i][j][k]; D+=a[last],P+=b[last]; if(j==1) return; dfs(last-1,j-1,k-(a[last]-b[last])); } int n,m; int main(){ scanf("%d%d",&n,&m); while(n && m){ cnt=0,D=0,P=0,u=0,num=0; memset(f,-1,sizeof(f)); memset(d,0,the sizeof (D)); F [ 0 ] [m * 20 is ] = 0 ; // negative as possible, so that a "translation array" for ( int I = . 1 ; I <= n-; I ++) Scanf ( " % D% D " , & A [I], & B [I]); for ( int I = . 1 ; I <= n-; ++ I) { for ( int J = M- . 1 ; J> = 0 ; - J) {// reverse cycle for ( int K = 0 ; K <= m * 40 ; ++ K) { D [I] [J+1][k+a[i]-b[i]]=d[i-1][j+1][k+a[i]-b[i]];//若不选i if(f[j][k]>=0){ if(f[j+1][k+a[i]-b[i]]<=f[j][k]+a[i]+b[i]){//选i f[j+1][k+a[i]-b[i]]=f[j][k]+a[i]+b[i]; d[i][j+1][k+a[i]-b[i]]=i; } } } } } for(int i=0;i<=m*40;++i){ if(f[m][i]<0) continue; if(pos>abs(i-20*m)){ pos=abs(i-20*m); u=i; num=f[m][i]; }else if(pos==abs(i-20*m)){ if(num<f[m][i]){ u=i; num=f[m][i]; } } } dfs(n,m,u); sort(p+1,p+1+cnt); printf("Jury #%d\nBest jury has value %d for prosecution and value %d for defence:\n ",++t,D,P); for(int i=1;i<=cnt;++i) cout<<p[i]<<" "; cout<<endl; scanf("%d%d",&n,&m); } return 0; }